Our aim is to find a condition for the existence of a hyperbolic
circle through the vertices of a hyperbolic triangle in terms of the lengths
of its sides.
We begin with a theorem which holds in both euclidean and hyperbolic
geometry, i.e. a theorem of neutral geometry.
In the theorem, we consider the size of angles, and assume that each
angle is taken in the range (0,π). That is to say, we take <XYZ as the
measure of the internal angle at vertex Y in ΔXYZ.
the angle theorem
For any triangle ΔABC, and any point P which is not on any
of the lines AB, BC, CA. Let <APB=γ, <BPC=α, <CPA=β, then
cos^{2}(α) + cos^{2}(β) + cos^{2}(γ) =
2cos(α)cos(β)cos(γ) + 1.
proof
Although this result has nice applications in euclidean geometry, we shall
concentrate on the hyperbolic case.
basic lemma
Suppose that ΔABC is a hyperbolic triangle
and that P is a point not on any of the hyperbolic lines AB, BC, CA.
Let d(A,B)=c, d(B,C)=a, d(C,A)=b, d(P,A)=r, d(P,B)=s, d(P,C)=t.
Then G(cosh(r),cosh(s),cosh(t),cosh(a),cosh(b),cosh(c)) = 0,
where G(R,S,T,X,Y,Z) is given by
(T^{2}1)(RSZ)^{2} +
(S^{2}1)(RTY)^{2} +
(R^{2}1)(TSX)^{2}
2(RSZ)(STX)(TRY)  (R^{2}1)(S^{2}1)(T^{2}1).
This is obtained from the angle theorem and the cosine rule
for the angles at P.
It is easy to see that G is quadratic in each of R, S, T.
A calculation shows that the coefficient of T^{2} is Z^{2}1.
Since we will have Z = cosh(c) > 1, G is quadratic in T.
Now observe that, if ΔABC has a hyperbolic circumcircle K(P,r), then
we can apply the lemma to the hyperbolic centre P. We then have
R = S = T = cosh(r), and the condition becomes
(R^{2}1){(R^{2}Z)^{2}+
(R^{2}Y)^{2}+
(R^{2}X)^{2}}2(R^{2}Z)(R^{2}X)(R^{2}Y)
(R^{2}1)^{3} = 0,
where X =cosh(a), Y = cosh(b), Z = cosh(c), R = cosh(r).
The left hand side is clearly a polynomial in R^{2}, and appears to involve
terms in R^{4} and R^{6}. If we multiply out and simplify, we get the equation
F(cosh(a),cosh(b),cosh(c))R^{2} = Δ^{2}(cosh(a),cosh(b),cosh(c)),
where
F(X,Y,Z) = 3+2(XY+YZ+ZX)2(X+Y+Z)(X^{2}+Y^{2}+Z^{2}), and
Δ^{2}(X,Y,Z) = 1+2XYZ(X^{2}+Y^{2}+Z^{2}).
This is the the circumradius equation
By expanding then simplifying we get the difference formula
Δ^{2}(X,Y,Z)F(X,Y,Z) = 2(X1)Y1)(Z1).
We have met Δ^{2} before, in connection with the trigonometry of hyperbolic
triangles. From that context, we know that it must be positive. We take
Δ as the positive root of this quantity.
It follows that, as R^{2} > 0, we must have F(cosh(a),cosh(b),cosh(c)) > 0.
We have arrived at the following
necessary condition for a hyperbolic circumcircle
If the hyperbolic triangle ΔABC has a hyperbolic circumcircle, then
(1) F(cosh(a),cosh(b),cosh(c)) > 0, and
(2) the hyperbolic radius r is determined by
cosh(r) = Δ(cosh(a),cosh(b),cosh(c))/F^{½}(cosh(a),cosh(b),cosh(c)).
The other hyperbolic functions of r are easily determined from cosh(r).
The manipulation is a little tedious, but we obtain
sinh^{2}(r) = 2(cosh(a)1)cosh(b)1)(cosh(c)1)/F(cosh(a),cosh(b),cosh(c)),
tanh^{2}(r) = 2(cosh(a)1)cosh(b)1)(cosh(c)1)/Δ^{2}(cosh(a),cosh(b),cosh(c)).
The formula for sinh(r) shows again that we must have F > 0.
The formula for tanh(r) gives an alternative formulation of the condition
Alternative necessary condition
If ΔABC has a hyperbolic circumcircle, then we must have
2(cosh(a)1)cosh(b)1)(cosh(c)1) < Δ^{2}(cosh(a),cosh(b),cosh(c)).
This is immediate as tanh(r) < 1.
If we observe that cosh(x)1 = 2sinh^{2}(½x), the formula for tanh(r) yields
tanh(r) = 4sinh(½a)sinh(½b)sinh(½c)/Δ(cosh(a),cosh(b),cosh(c))
The removal of the need to take a root is an illusion  Δ involves a root.
There are various ways to rewrite the formulae for r. Some are given as
a digression.
Our main purpose is to show that either of our conditions are also sufficient
to guarantee the existence of a circumcircle. This is not so easy. We give a
proof in a separate page.
